If $(a_{1} + i b_{1}) (a_{2} + i b_{2}) . . . . . . . (a_{n} + i b_{n}) = A + i B$ then $({a_{1}}^{2} + {b_{1}}^{2}) ({a_{2}}^{2} + {b_{2}}^{2}) . . . . . . . . . ({a_{n}}^{2} + {b_{n}}^{2})$ equals to :

1

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A^{2} + B^{2}

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A + B

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\frac{1}{A^{2}} + \frac{1}{B^{2}}

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Solution

The correct option is B $A^{2} + B^{2}$
$(a_{1} + {i b}_{1}) (a_{2} + {i b}_{2}) . . . . . . (a_{n} + {i b}_{n}) = A + i B ⟶ (1)$
Taking conjugates,
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (a_{1} + {i b}_{1}) (a_{2} + {i b}_{2}) . . . . . . (a_{n} + {i b}_{n}) = ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A + i B ⟶ (2)$
We know that,
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} z_{2} z_{3} . . . . . . z_{n} =_{1}_{2} . . . . . ._{n} ⟶ (3)$
Using property $(3)$ in $(2)$ we can write,
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (a_{1} + {i b}_{1}) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (a_{2} + {i b}_{2}) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (a_{3} + {i b}_{3}) . . . . . . ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (a_{n} + {i b}_{n}) = ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A + i B$
$(a_{1} - {i b}_{1}) (a_{2} - {i b}_{2}) (a_{3} - {i b}_{3}) . . . . . . (a_{n} - {i b}_{n}) = A - i B ⟶ (4)$
Multiplying $(4)$ & $(1)$ we get
$(a_{1}^{2} + b_{1}^{2}) (a_{2}^{2} + b_{2}^{2}) . . . . . . ({a n}^{2} + b_{n}^{2}) = (A + i B) (A - i B) = A^{2} + B^{2}$

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